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a: Số số hạng là:

(2n-2):2+1=n(số)

Theo đề, ta có:

\(\left(2n+2\right)\cdot\dfrac{n}{2}=210\)

\(\Leftrightarrow n\left(n+1\right)=210\)

\(\Leftrightarrow n=14\)

1 tháng 11 2021

Ta có : 2 + 4 + 6 + ... + 2(n - 1) + 2n = 210

<=> 2[1 + 2 + 3 + ... + (n - 1) + n] = 210

<=> 1 + 2 + 3 + ... + n = 105

<=> [(n - 1) : 1 + 1)(n + 1) : 2 = 105

<=> n(n + 1) = 210

<=> n(n + 1) = 14.15

=> n = 14

Vậy n = 14

b) Ta có : 1 + 3 + 5 + ... + (2n - 1) = 225

<=> [(2n - 1 - 1) : 2 + 1](2n - 1 + 1) : 2 = 225

<=> n2 = 225

<=> n2 = 152

<=> n = 15

Vậy n = 15

1 tháng 11 2021
210 = 2 + 4 + 6 + ...+ 2n = n(2 + 2n)/2 = n(1 + n) = n^2 + n n^2 + n - 210 = 0 => n = -15 (loại); n = 14 225 = 1 +3 + 5 +...+ (2n + 1) = (n + 1)(2n + 1 + 1)/2 = (n + 1)^2 n + 1 = 15 n = 14
8 tháng 7 2023

a) \(2^n=8\)

\(\Rightarrow2^n=2^3\)

\(\Rightarrow n=3\)

b) \(5^{n+1}=125\)

\(\Rightarrow5^{n+1}=5^3\)

\(\Rightarrow n+1=3\)

\(\Rightarrow n=3-1=2\)

c) Mình không rõ đề:

d) \(2\cdot7^{n-1}+3=101\)

\(\Rightarrow2\cdot7^{n-1}=101-3\)

\(\Rightarrow2\cdot7^{n-1}=98\)

\(\Rightarrow7^{n-1}=\dfrac{98}{2}\)

\(\Rightarrow7^{n-1}=49\)

\(\Rightarrow7^{n-1}=7^2\)

\(\Rightarrow n-1=2\)

\(\Rightarrow n=1+2=3\)

e) \(3\cdot5^{2n+1}-6^2=339\)

\(\Rightarrow3\cdot5^{2n+1}=339+36\)

\(\Rightarrow3\cdot5^{2n+1}=375\)

\(\Rightarrow5^{2n+1}=125\)

\(\Rightarrow5^{2n+1}=5^3\)

\(\Rightarrow2n+1=3\)

\(\Rightarrow2n=2\)

\(\Rightarrow n=\dfrac{2}{2}=1\)

20 tháng 10 2023

Mình mẫu đầu với cuối nhé:

a)  Đặt \(ƯCLN\left(3n+4,3n+7\right)=d\)  

\(\Rightarrow\left\{{}\begin{matrix}3n+4⋮d\\3n+7⋮d\end{matrix}\right.\)

\(\Rightarrow\left(3n+7\right)-\left(3n+4\right)⋮d\)

\(\Rightarrow3⋮d\)

 \(\Rightarrow d\in\left\{1,3\right\}\)

Nhưng do \(3n+4,3n+7⋮̸3\) nên \(d\ne3\Rightarrow d=1\)

Vậy \(ƯCLN\left(3n+4,3n+7\right)=1\) hay \(3n+4,3n+7\) nguyên tố cùng nhau.

 e) \(ƯCLN\left(2n+3,3n+5\right)=d\)

 \(\Rightarrow\left\{{}\begin{matrix}2n+3⋮d\\3n+5⋮d\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}6n+9⋮d\\6n+10⋮d\end{matrix}\right.\)

\(\Rightarrow\left(6n+10\right)-\left(6n+9\right)⋮d\)

\(\Rightarrow1⋮d\) \(\Rightarrow d=1\)

Vậy \(ƯCLN\left(2n+3,3n+5\right)=1\), ta có đpcm.

8 tháng 8 2023

a, Ta có : \(\text{n + 5 = (n - 1)+6}\)

Vì \(\text{(n-1) ⋮ n-1}\)

Nên để \(\text{n+5 ⋮ n-1}\) `n-1`

Thì \(\text{6 ⋮ n-1}\) 

\(\Rightarrow\) \(\text{n - 1 ∈ Ư(6)}\)

\(\Rightarrow\) \(\text{n - 1 ∈}\) \(\left\{\text{±1;±2;±3;±6}\right\}\)

\(\Rightarrow\) \(\text{n ∈}\) \(\left\{\text{0;-1;-2;-5;2;3;4;7}\right\}\) \(\text{( TM )}\)

\(\text{________________________________________________________}\)

b, Ta có : \(\text{2n-4 = (2n+4)- 8 = 2(n+2) - 8}\)

Vì \(\text{2(n+2) ⋮ n+2}\)

Nên để \(\text{2n-4 ⋮ n+2}\)

Thì \(\text{8 ⋮ n+2}\)

\(\Rightarrow\) \(\text{n + 2 ∈ Ư(8)}\)

\(\Rightarrow\) \(\text{n + 2 ∈}\) \(\left\{\text{±1;±2;±4;±8}\right\}\)

\(\Rightarrow\) \(\text{n ∈}\) \(\left\{\text{-3;-4;-6;-10;-1;0;2;6}\right\}\) ( TM )

\(\text{_________________________________________________________________ }\)

c, Ta có :\(\text{ 6n + 4 = (6n + 3) +1 = 3(2n+1) + 1}\)

Vì \(\text{3(2n+1) ⋮ 2n+1}\)

Nên để\(\text{ 6n+4 ⋮ 2n+1}\)

Thì \(\text{1 ⋮ 2n+1}\)

\(\Rightarrow\) \(\text{2n + 1 ∈ Ư(1)}\)

\(\Rightarrow\) \(\text{2n + 1 ∈}\) \(\left\{\text{±1}\right\}\)

\(\Rightarrow\) \(\text{2n ∈}\) \(\left\{\text{-2;0}\right\}\)

\(\Rightarrow\) \(\text{n ∈}\) \(\left\{\text{-1;0}\right\}\) ( TM )

\(\text{_______________________________________}\)

Ta có : \(\text{3 - 2n = -( 2n - 3 ) = -( 2n + 2 ) + 5 = -2( n+1)+5}\)

Vì \(\text{-2(n+1) ⋮ n+1}\)

Nên để \(\text{3-2n ⋮ n+1}\)

Thì\(\text{ 5 ⋮ n + 1}\)

\(\Rightarrow\) \(\text{n + 1 ∈}\) \(\left\{\text{±1;±5}\right\}\)

\(\Rightarrow\) \(\text{n ∈}\) \(\text{-2;-6;0;4}\) ( TM )

 

19 tháng 8 2017

1 . tìm giá trị x 

\(\left(x+1\right)+\left(x+4\right)+....+\left(x+28\right)=115.\)

\(\Rightarrow\left(x+x+x+....x\right)+\left(1+4+..+28\right)=115\)

\(\Rightarrow10x+\left(28+1\right).10:2=115\)

\(\Rightarrow10x+145=115\)

\(\Rightarrow10x=115-145=-30\)

\(\Rightarrow x=-30:10=-3\)

21 tháng 11 2016

Bai 1

so so hang la: [(28+x)-(x+1)]/3+1= 10 so hang

tong =[(x+1)+(x+28)]*10/2=(2x+29)*10/2=115

(2x+29)*5=115

2x+29=115/5=23

2x=23-29=-6

x=-3

26 tháng 10 2020

Ta có: \(\frac{2n+6}{2n-1}=\frac{2n-1+7}{2n-1}=1+\frac{7}{2n-1}\)

để \(2n-6⋮2n-1\) thì \(7⋮2n-1\)

hay 2n -1 \(\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

xét bảng 

2n-11-17-7
2n208-6
n104-3

vậy........

a, \(\dfrac{15}{n-1}\); n∈Z

\(\dfrac{15\left(n-1\right)}{n-1}=\dfrac{15n-15}{n-1}\)

=> Ư(15)={\(\pm1;\pm3;\pm5;\pm15\)}

n-1 -15 -5 -3 -1 1 3 5 15
n -14 -4 -2 0 2 4 6 16
Đánh giá  t/mãn t/mãn t/mãn t/mãn t/mãn t/mãn t/mãn t/mãn

 

Vậy n∈{-14;-4;-2;0;2;4;6;16}

b, \(\dfrac{-21}{n+3}\)  n∈Z

\(\dfrac{-21\left(n+3\right)}{n+3}=\dfrac{\left(-21n-63\right)}{n+3}\)

Ư(63)={±1;±3;±7;±9;±21;±63}

n+3 -63 -21 -9 -7 -3 -1 1 3 7 9 21 63
n -66 -24 -12 -10 -6 -4 -2 0 4 6 18 60
Đ/gia t/mãn t/mãn t/mãn t/mãn t/mãn t/mãn t/mãn t/mãn t/mãn t/mãn t/mãn t/mãn

 

Vậy n∈{-66;-24;-12;-10;-6;-4;-2;0;4;6;18;60}

 

 

\(\dfrac{2n+7}{n-2};n\inℤ\\ \Rightarrow\dfrac{\left(2n-4\right)+7+2}{n-2}=\dfrac{2\left(n-2\right)+9}{n-2}=2+\dfrac{9}{n-2}\)

\(\LeftrightarrowƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)

Ta có bảng sau:

n-2 -9 -3 -1 1 3 9
n -7 -1 1 3 5 11
Đ/gia t/mãn t/mãn t/mãn t/mãn t/mãn t/mãn

 

Vậy n={-7;-1;1;3;5;11}

 

a)Gọi ƯCLN (\(n+3;2n+5\))=d

\(\Rightarrow\left\{{}\begin{matrix}\left(n+3\right)⋮d\Rightarrow2\left(n+3\right)⋮d\Rightarrow\left(2n+6\right)⋮d\\\left(2n+5\right)⋮d\end{matrix}\right.\)

\(\Rightarrow\left(2n+6\right)-\left(2n+5\right)⋮d\Rightarrow1⋮d\Rightarrow d=1\)

⇒ƯCLN (\(n+3;2n+5\))=1

\(\Rightarrow\frac{n+3}{2n+5}\)là phân số tối giản(đpcm)

b)Gọi ƯCLN (\(2n+9;3n+14\))=d

\(\Rightarrow\left\{{}\begin{matrix}\left(2n+9\right)⋮d\Rightarrow3\left(2n+9\right)⋮d\Rightarrow\left(6n+27\right)⋮d\\\left(3n+14\right)⋮d\Rightarrow2\left(3n+14\right)⋮d\Rightarrow\left(6n+28\right)⋮d\end{matrix}\right.\)

\(\Rightarrow\left(6n+28\right)-\left(6n+27\right)⋮d\Rightarrow1⋮d\Rightarrow d=1\)

⇒ƯCLN (\(2n+9;3n+14\))=1

\(\Rightarrow\frac{2n+9}{3n+14}\) là phân số tối giản.(đpcm)

c)Gọi ƯCLN(\(6n+11;2n+5\))=d

\(\Rightarrow\left\{{}\begin{matrix}\left(6n+11\right)⋮d\\\left(2n+5\right)⋮d\Rightarrow3\left(2n+5\right)⋮d\Rightarrow\left(6n+15\right)⋮d\end{matrix}\right.\)

\(\Rightarrow\left(6n+15\right)-\left(6n+11\right)⋮d\)

\(\Rightarrow4⋮d\)

\(\left(6n+15\right);\left(6n+11\right)⋮̸2\)

\(\Rightarrow d=1\)

⇒ƯCLN(\(6n+11;2n+5\))=1

\(\Rightarrow\frac{6n+11}{2n+5}\)là phân số tối giản (đpcm)

d)Gọi ƯCLN(\(12n+1;30n+2\))=d

\(\Rightarrow\left\{{}\begin{matrix}\left(12n+1\right)⋮d\Rightarrow5\left(12n+1\right)⋮d\Rightarrow\left(60n+5\right)⋮d\\\left(30n+2\right)⋮d\Rightarrow2\left(30n+2\right)⋮d\Rightarrow\left(60n+4\right)⋮d\end{matrix}\right.\)

\(\Rightarrow\left(60n+5\right)-\left(60n+4\right)⋮d\)

\(\Rightarrow1⋮d\Rightarrow d=1\)

⇒ƯCLN(\(12n+1;30n+2\))=1

\(\Rightarrow\frac{12n+1}{30n+2}\) là phân số tối giản (đpcm)

e)Gọi ƯCLN(\(21n+4;14n+3\))=d

\(\Rightarrow\left\{{}\begin{matrix}\left(21n+4\right)⋮d\Rightarrow2\left(21n+4\right)⋮d\Rightarrow\left(42n+8\right)⋮d\\\left(14n+3\right)⋮d\Rightarrow3\left(14n+3\right)⋮d\Rightarrow\left(42n+9\right)⋮d\end{matrix}\right.\)

\(\Rightarrow\left(42n+9\right)-\left(42n+8\right)⋮d\Rightarrow1⋮d\Rightarrow d=1\)

⇒ƯCLN(\(21n+4;14n+3\))=1

\(\Rightarrow\frac{21n+4}{14n+3}\)là phân số tối giản (đpcm)

f) Gọi ƯCLN(\(2n+3;n+2\))=d

\(\Rightarrow\left\{{}\begin{matrix}\left(2n+3\right)⋮d\\\left(n+2\right)⋮d\Rightarrow2\left(n+2\right)⋮d\Rightarrow\left(2n+4\right)⋮d\end{matrix}\right.\)

\(\Rightarrow\left(2n+4\right)-\left(2n+3\right)⋮d\Rightarrow1⋮d\Rightarrow d=1\)

⇒ƯCLN(\(2n+3;n+2\))=1

\(\Rightarrow\frac{2n+3}{n+2}\)là phân số tối giản (đpcm)
g) Gọi ƯCLN(\(n+1;3n+2\))=d

\(\Rightarrow\left\{{}\begin{matrix}\left(n+1\right)⋮d\Rightarrow3\left(n+1\right)⋮d\Rightarrow\left(3n+3\right)⋮d\\\left(3n+2\right)⋮d\end{matrix}\right.\)

\(\Rightarrow\left(3n+3\right)-\left(3n+2\right)⋮d\Rightarrow1⋮d\Rightarrow d=1\)

⇒ƯCLN(\(n+1;3n+2\))=1

\(\Rightarrow\frac{n+1}{3n+2}\) là phân số tối giản (đpcm)